To find the 35th term of the Fibonacci sequence using Binet's Formula, we use the following equation:

$F(n) = \frac{ \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n }{\sqrt{5}}$

Where:

• $F(n)$ is the nth Fibonacci number,
• $n$ is the position in the sequence (in this case, $n = 35$).

Let’s calculate the 35th term using this formula. I'll simplify the terms and compute the result.

$F(35) = \frac{ \left( \frac{1 + \sqrt{5}}{2} \right)^{35} - \left( \frac{1 - \sqrt{5}}{2} \right)^{35} }{\sqrt{5}}$

Approximate calculation:

1. $\frac{1 + \sqrt{5}}{2} \approx 1.618$
2. $\frac{1 - \sqrt{5}}{2} \approx -0.618$

Now, using these values and computing:

$F(35) \approx \frac{1.618^{35} - (-0.618)^{35}}{\sqrt{5}}$

Let me compute the value.The 35th term of the Fibonacci sequence is approximately $9,227,465$.

Thus, the correct response is C.

Would you like more details on how this was calculated, or do you have any questions?

Here are 5 related questions to consider:

1. How does Binet's formula approximate Fibonacci numbers?
2. What is the significance of the golden ratio $\phi$ in Fibonacci calculations?
3. Can you use Binet’s formula for any position in the Fibonacci sequence?
4. How do rounding errors affect the precision of Binet's formula for large $n$?
5. What are other methods to calculate large Fibonacci numbers besides Binet's formula?

Tip: Binet's formula is an approximation, but it becomes highly accurate for larger $n$ because the negative term $\psi^n$ approaches 0.